Mirage

Keith,

The key here is the light from the target travels the path once. The light from the laser travels the path twice, doubling the displacement and making it visible relative to the crosshairs.

Jim
And this is true because the image of the crosshairs does not make the 100 yard trip at all.
 
Keith

A possible simple demonstration occurred to me last night.

Mirage is caused by a difference in refractive index in the image path changing the direction of the light. Lets take this to an extreme for demonstration purposes and use water.

Ever been bowfishing? If you aim at where the fish "looks" to be you'll miss. Stick a pole in the water and it appears to "bend" even though it remains perfectly straight. This pole represents the actual arrow path. So you have to aim high to hit low.

Lets put a target at the bottom of an empty pool - shine a laser on the bullseye at some angle other than vertical and center the crosshairs on the bullseye.

Once we've done that - fill the pool with water.

The water will refract the laser, causing it to no longer hit the target in the center - it will hit lower now due to this bending.

The target will also appear to have moved in the crosshairs to a point below where it really is.

So now we have the image of the bullseye located 1/2 way between the crosshairs and the image of the laser spot.

Knowing this relationship will allow a shooter to be able to determine just what the error due to mirage is going to be. So he can compensate for it.
 
Keith,

The key here is the light from the target travels the path once. The light from the laser travels the path twice, doubling the displacement and making it visible relative to the crosshairs.

Jim

Jim,
Thanks for the explanation. I see where the problem lies. On the way back the laser is refracted not in the same direction to double the displacement, but in the opposite direction to return the displacement to zero. For example, for a laser striking an air/water interface at 45 degrees, it is refracted to angle theta2 given by sin(theta2) = sin(theta1)*n1/n2 (Snell's law), where theta1 is 45 degrees and the indices of refraction are for air n1 = 1 and for water n2 = 1.33. This gives theta2 = 32 degrees. On the way back, the laser strikes the WATER/AIR (the interface is now reversed) interface at 32 degrees and is refracted to 45 degrees. Thus the laser follows the same path on the way back as the way out.

Cheers,
Keith
 
Keith

A possible simple demonstration occurred to me last night.

Mirage is caused by a difference in refractive index in the image path changing the direction of the light. Lets take this to an extreme for demonstration purposes and use water.

Ever been bowfishing? If you aim at where the fish "looks" to be you'll miss. Stick a pole in the water and it appears to "bend" even though it remains perfectly straight. This pole represents the actual arrow path. So you have to aim high to hit low.

Lets put a target at the bottom of an empty pool - shine a laser on the bullseye at some angle other than vertical and center the crosshairs on the bullseye.

Once we've done that - fill the pool with water.

The water will refract the laser, causing it to no longer hit the target in the center - it will hit lower now due to this bending.

OK to here.

The target will also appear to have moved in the crosshairs to a point below where it really is.

Because the image follows the same path as the laser, the bull moves up.

So now we have the image of the bullseye located 1/2 way between the crosshairs and the image of the laser spot.

There is only one light path and both the laser and the image of the target follow it. Otherwise Snell's law is violated.

Actually, this would be a good demonstration that could resolve our disagreement based on theoretical grounds. Tape a laser pointer to a camera on a tripod, and follow the directions you have given. Here is a picture of a laser being refracted into water:
http://www.warren-wilson.edu/~physics/PhysPhotOfWeek/20050401LaserRefraction/index.shtml

Keith
 
Keith,

take another look at the picture you posted the link to. I have no idea what the depth of the water was, but it is beautiful example of why the laser does NOT hit the same point. The entry angles and the exit angles are EXACTLY the same, BUT....the reflected image is MOVED because of the refractive effect of the water. If you were to shoot based on the laser line of exit, you would a distance off based on the separation of the 2 lines of light. That is not an exact equivalent of mirage, since air is not as dense as water, but relativistically, it is the same. There is no way to get a real picture of the target with the equipment we are using. We are always looking at bended light, even when there is no apparent mirage. the key is to understand that, believe it, and use it to our advantage. the really good shooters do it, even if they do not know it.

OK to here.



Because the image follows the same path as the laser, the bull moves up.



There is only one light path and both the laser and the image of the target follow it. Otherwise Snell's law is violated.

Actually, this would be a good demonstration that could resolve our disagreement based on theoretical grounds. Tape a laser pointer to a camera on a tripod, and follow the directions you have given. Here is a picture of a laser being refracted into water:
http://www.warren-wilson.edu/~physics/PhysPhotOfWeek/20050401LaserRefraction/index.shtml

Keith
 
Keith,

take another look at the picture you posted the link to. I have no idea what the depth of the water was, but it is beautiful example of why the laser does NOT hit the same point. The entry angles and the exit angles are EXACTLY the same, BUT....the reflected image is MOVED because of the refractive effect of the water. If you were to shoot based on the laser line of exit, you would a distance off based on the separation of the 2 lines of light. That is not an exact equivalent of mirage, since air is not as dense as water, but relativistically, it is the same. There is no way to get a real picture of the target with the equipment we are using. We are always looking at bended light, even when there is no apparent mirage. the key is to understand that, believe it, and use it to our advantage. the really good shooters do it, even if they do not know it.

David,
The question isn't whether the laser is refracted, because the picture shows that it obviously is. The question is whether one can tell where the bullseye really is from an image obtained along the same refracted path as the laser.

I wish I knew how to shoot mirage. I had some costly vertical flyers last weekend that I am thinking were caused by mirage.

Cheers,
Keith
 
Lasers aside for a moment, if one had a really sturdy bench, or a specially built stand that was heavy and rigid, could one not mount a rifle scope on that stand, and align it with a fixed target, that was not changed throughout the day. If the scope was aligned to the center of the target just at dawn, before the sun had heated the ground, it could be used as a reference to show the effects of mirage. If you wanted to go high tech. you could put a video camera behind the reference scope, and wire it to a small monitor that was mounted so as to be convenient for the shooter to see as he was shooting.
 
Keith
You and Boyd are both correct. I'm big enough to admit I made an error. (Mainly because I found batteries for my laser and tried it out, but still...:). Congratulations, and thanks for your perseverance. Unlike some, I can change my mind in the face of hard facts.

One cannot use a riflescope mounted laser to illustrate mirage. For exactly the reasons you presented. It's a shame, as it would have been handy as all get out....But I digress.

However, Boyd's idea of mounting a scope (or a laser) to the bench, allowing it to remain fixed on a given target for the duration of the match, would indeed work. As the mirage shifted the light path the bull would appear to move under the crosshairs and/or laser spot. This reference target could be checked in much the same way as wind flags are checked as part of reading the conditions. And mirage becomes one of the more readable conditions....Even in a boil.

In fact, it wouldn't even have to be set up in a zero mirage condition, since it relative change that we are concerned about. Turret adjustments can correct for the baseline condition.
 
Lasers aside for a moment, if one had a really sturdy bench, or a specially built stand that was heavy and rigid, could one not mount a rifle scope on that stand, and align it with a fixed target, that was not changed throughout the day. If the scope was aligned to the center of the target just at dawn, before the sun had heated the ground, it could be used as a reference to show the effects of mirage. If you wanted to go high tech. you could put a video camera behind the reference scope, and wire it to a small monitor that was mounted so as to be convenient for the shooter to see as he was shooting.
Boyd,

Your idea is exactly what I’ve been using for the last several years. My reference scope sits in a steel v-block atop my shooting platform which is composed of a nearly 2 Ton block of concrete. But, it’s not an original idea on my part because I stole it from Harold Vaughn.
I’ve never utilized it outside of my tunnel, but if I raised my overhead door it’s possible for me to sight in across several thousand yards once all the crops are harvested. That might be both fun and educational for me since I don’t ever have to deal with any wind or much mirage in my tunnel and consequently I’m a real novice at it.

Landy
 
Perhaps you could post targets with rings a that were scaled for visibility at the ranges that they were posted and let us know what you observe. I think that It might be interesting.
 
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